Show that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|\ dx$ diverges

Question

I read that the improper Riemann integral $$\int_0^1 \Bigg|\frac{1}{x}\sin\frac{1}{x}\Bigg|\ dx$$ diverges.

I have tried comparison criteria for $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx$, but I cannot find a function $f$ with a divergent integral such that $0\leq f(x)\leq|\frac{1}{x}\sin\frac{1}{x}|$. I also notice, by using a change of variable, that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx=\int_1^\infty |\frac{1}{x}\sin x|dx$, but I have not found a use of this equality to prove the divergence of the integral. I have also tried to think about some use of complex analysis, but found none useful to prove the desired result. Coud anybody give a proof of this divergence? $\infty$ thanks!

Answer 1

I also notice, by using a change of variable, that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx=\int_1^\infty |\frac{1}{x}\sin x|dx$, but I have not found a use of this equality to prove the divergence of the integral.

Subdivide $(1,\infty)$ into an infinite number of intervals of the form $\Big(k\pi,~(k+1)\pi\Big)$, and notice that for each of them, their graphic is tangent to a hyperbola, and that the area of each slice is greater than that of a triangle of height $\dfrac1{\pi~\bigg(k+\dfrac12\bigg)}$, due to the convexity/concavity of the sine function. This shows that the value of our integral is greater than a constant multiple of the harmonic series, which is known to be divergent.

Answer 2

Hint: Calculate the integral only for those intervals where $\sin 1/x ≥ 1/2$.

Answer 3

Performing the changes of variables $x= 1/u$ you get

$$\int_0^1 \bigg|\frac{1}{x}\sin\frac{1}{x}\bigg|\ dx=\int_1^\infty \bigg|\frac{\sin x}{x}\ \bigg| dx\ge \int_π^\infty \Bigg|\frac{\sin x}{x}\ \bigg|\\= \lim_{n\to \infty}\int_π^{nπ} \bigg|\frac{\sin x}{x}\ \bigg|dx= \sum_{n=1}^{\infty}\int_{nπ}^{(n+1)π} \bigg|\frac{\sin x}{x}\ \bigg|dx\\ \ge \sum_{n=1}^{\infty}\frac{1}{ (n+1)π} \int_{nπ}^{(n+1)π} |\sin x|dx =\color{red}{\sum_{n=1}^{\infty}\frac{2}{ (n+1)π}=\infty} $$

indeed for all $n$ $$\color{blue}{\int_{nπ}^{(n+1)π} |\sin x|dx=2}$$

Also see here: Is this proof correct? Divergence of $\int_{1}^{\infty} \left| \frac{\sin x}{x} \right| \, \mathrm{d}x $